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E - Pavel and barbecue CodeForces - 760C

岳泉
2023-12-01

Pavel cooks barbecue. There are n skewers, they lay on a brazier in a row, each on one of n positions. Pavel wants each skewer to be cooked some time in every of n positions in two directions: in the one it was directed originally and in the reversed direction.

Pavel has a plan: a permutation p and a sequence b1, b2, …, bn, consisting of zeros and ones. Each second Pavel move skewer on position i to position pi, and if bi equals 1 then he reverses it. So he hope that every skewer will visit every position in both directions.

Unfortunately, not every pair of permutation p and sequence b suits Pavel. What is the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements? Note that after changing the permutation should remain a permutation as well.

There is no problem for Pavel, if some skewer visits some of the placements several times before he ends to cook. In other words, a permutation p and a sequence b suit him if there is an integer k (k ≥ 2n), so that after k seconds each skewer visits each of the 2n placements.

It can be shown that some suitable pair of permutation p and sequence b exists for any n.

Input
The first line contain the integer n (1 ≤ n ≤ 2·105) — the number of skewers.

The second line contains a sequence of integers p1, p2, …, pn (1 ≤ pi ≤ n) — the permutation, according to which Pavel wants to move the skewers.

The third line contains a sequence b1, b2, …, bn consisting of zeros and ones, according to which Pavel wants to reverse the skewers.

Output
Print single integer — the minimum total number of elements in the given permutation p and the given sequence b he needs to change so that every skewer will visit each of 2n placements.

Example
Input
4
4 3 2 1
0 1 1 1
Output
2
Input
3
2 3 1
0 0 0
Output
1
Note
In the first example Pavel can change the permutation to 4, 3, 1, 2.

In the second example Pavel can change any element of b to 1.

#include<iostream>
#include<cstring>
using namespace std;
#define LL long long int
#define INF 0x3f3f3f3f
const int maxn = 3e5 + 10;
int p[maxn];
int b[maxn];
int vis[maxn];
void dfs(int x)
{
    vis[x] = 1;
    if (vis[p[x]] == 0)
    {
        dfs(p[x]);
    }
}
int main()
{
    int n;
    while (cin >> n)
    {
        int ans = 0; int tme = 0 ;
        memset(vis, 0, sizeof(vis));
        for (int i = 1; i <= n; i++)
        {
            cin >> p[i];
        }
        for (int i = 1; i <= n; i++)
        {
            cin >> b[i];
            ans += b[i];
        }
        if (ans % 2 == 0)
        {
            ans = 1;
        }
        else
            ans = 0;
        tme = 0;
        for (int i = 1; i <= n; i++)
        {
            if (vis[i] == 0)
            {
                tme++;
                dfs(i);
            }
        }
        if (tme == 1)
            tme = 0;
        cout << tme + ans << endl;
    }
    return 0;
}
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